Let $R$ be the radius of the given cone, $r$ the radius of the small cone, $h$ be the height of the frustum and $h_1$ be the height of the small cone.

In the figure,$△ONC\sim △OMA$
$\therefore \frac{ON}{OM}=\frac{NC}{MA}$    [Sides of similar triangles are proportional]
$\Rightarrow \frac{h_1}{40}=\frac{r}{R}$ 
$\Rightarrow h_1=\left(\frac{r}{R}\right)\times 40$ .......(i)

We are given that $\frac{Volumeofsmallcone}{volumeofgivencone}=\frac{1}{64}$ 

$\Rightarrow \frac{\frac{1}{3}\pi r^2\times h_1}{\frac{1}{3}\pi R^2\times 40}=\frac{1}{64}$

$\Rightarrow \frac{r^2}{R^2}\times \frac{1}{40}\times \left[\left(\frac{r}{R}\right)40\right]=\frac{1}{64}$       [ By (i) ] 

$\Rightarrow {\left(\frac{r}{R}\right)}^3=\frac{1}{64}={\left(\frac{1}{4}\right)}^3$
$\Rightarrow$  $\frac{r}{R}=\frac{1}{4}$    ...... (ii)

From (i) and (ii) $h_1=\frac{1}{4}\times 40=10cm$

$\therefore ,h=40-h_1=(40-10)cm$

$\Rightarrow h=30cm$

Hence, the section is made at a height of 30 cm above the base of the cone  .